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On the expression of a number in the form ax2 + by 2 + cz 2 + du2 Proceedings of the Cambridge Philosophical Society, XIX, 1917, 11 – 21 1. It is well known that all positive integers can be expressed as the sum of four squares. This naturally suggests the question: For what positive integral values of a, b, c, d, can all positive integers be expressed in the form ax2 + by 2 + cz 2 + du2 ? (1.1) I prove in this paper that there are only 55 sets of values of a, b, c, d for which this is true. The more general problem of finding all sets of values of a, b, c, d for which all integers with a finite number of exceptions can be expressed in the form (1.1), is much more difficult and interesting. I have considered only very special cases of this problem, with two variables instead of four; namely, the cases in which (1.1) has one of the special forms a(x2 + y 2 + z 2 ) + bu2 , (1.2) a(x2 + y 2 ) + b(z 2 + u2 ). (1.3) and These two cases are comparatively easy to discuss. In this paper I give the discussion of (1.2) only, reserving that of (1.3) for another paper. 2. Let us begin with the first problem. We can suppose, without loss of generality, that a ≤ b ≤ c ≤ d. (2.1) if a > 1, then 1 cannot be expressed in the form (1.1); and so a=1 (2.2) 1 ≤ b ≤ 2. (2.3) If b > 2, then 2 is an exception; and so We have therefore only to consider the two cases in which (1.1) has one or other of the forms x2 + y 2 + cz 2 + du2 , x2 + 2y 2 + cz 2 + du2 . In the first case, If c > 3, then 3 is an exception; and so 1 ≤ c ≤ 3. (2.31) Paper 20 218 In the second case, if c > 5, then 5 is an exception; and so 2 ≤ c ≤ 5. (2.32) We can now distinguish 7 possible cases. (2.41) x2 + y 2 + z 2 + du2 . If d > 7, 7 is an exception; and so 1 ≤ d ≤ 7. (2.42) (2.411) x2 + y 2 + 2z 2 + du2 . If d > 14, 14 is an exception; and so 2 ≤ d ≤ 14. (2.43) (2.421) x2 + y 2 + 3z 2 + du2 . If d > 6, 6 is an exception; and so 3 ≤ d ≤ 6. (2.44) (2.431) x2 + 2y 2 + 2z 2 + du2 . If d > 7, 7 is an exception; and so 2 ≤ d ≤ 7. (2.45) (2.441) x2 + 2y 2 + 3z 2 + du2 . If d > 10, 10 is an exception; and so 3 ≤ d ≤ 10. (2.46) (2.451) x2 + 2y 2 + 4z 2 + du2 . If d > 14, 14 is an exception; and so 4 ≤ d ≤ 14. (2.47) (2.461) x2 + 2y 2 + 5z 2 + du2 . If d > 10, 10 is an exception; and so 5 ≤ d ≤ 10. (2.471) We have thus eliminated all possible sets of values of a, b, c, d, except the following 55∗ : L. E. Dickson (Bulletin of the Ameraican Math. Soc. [vol XXXIII (1927), pp. 63 – 70]) has pointed out that Ramanujan has overlooked the fact that (1, 2, 5, 5) does not represent 15. Consequently, there are only 54 forms. ∗ On the expression of a number in the form ax2 + by 2 + cz 2 + du2 1, 1, 1, 1 1, 2, 3, 5 1, 2, 4, 8 1, 1, 1, 2 1, 2, 4, 5 1, 2, 5, 8 1, 1, 2, 2 1, 2, 5, 5 1, 1, 2, 9 1, 2, 2, 2 1, 1, 1, 6 1, 2, 3, 9 1, 1, 1, 3 1, 1, 2, 6 1, 2, 4, 9 1, 1, 2, 3 1, 2, 2, 6 1, 2, 5, 9 1, 2, 2, 3 1, 1, 3, 6 1, 1, 2, 10 1, 1, 3, 3 1, 2, 3, 6 1, 2, 3, 10 1, 2, 3, 3 1, 2, 4, 6 1, 2, 4, 10 1, 1, 1, 4 1, 2, 5, 6 1, 2, 5, 10 1, 1, 2, 4 1, 1, 1, 7 1, 1, 2, 11 1, 2, 2, 4 1, 1, 2, 7 1, 2, 4, 11 1, 1, 3, 4 1, 2, 2, 7 1, 1, 2, 12 1, 2, 3, 4 1, 2, 3, 7 1, 2, 4, 12 1, 2, 4, 4 1, 2, 4, 7 1, 1, 2, 13 1, 1, 1, 5 1, 2, 5, 7 1, 2, 4, 13 1, 1, 2, 5 1, 1, 2, 8 1, 1, 2, 14 1, 2, 2, 5 1, 2, 3, 8 1, 2, 4, 14 1, 1, 1, 2 1, 1, 2, 4 1, 2, 4, 8 1, 1, 2, 2 1, 2, 2, 4 1, 1, 3, 3 1, 2, 2, 2 1, 2, 4, 4 1, 2, 3, 6 1, 1, 1, 4 1, 1, 2, 8 1, 2, 5, 10 1, 1, 3, 5 Of these 55 forms, the 12 forms have been already considered by Liouville and Pepin∗ . 3. I shall now prove that all integers can be expressed in each of the 55 forms. In order to prove this we shall consider the seven cases (2.41)–(2.47) of the previous section separately. We shall require the following results concerning ternary quadratic arithmetical forms. ∗ There are a large number of short notes by Liouville in Vols. V–VIII of the second series of his Journal. See also Pepin, ibid., Ser.4, Vol. VI, pp.1 – 67. The object of the work of Liouville and Pepin is rather different from mine, viz., to determine, in a number of special cases, explicit formulæ for the number of representations, in terms of other arithmetical functions. 219 Paper 20 220 The necessary and sufficient condition that a number cannot be expressed in the form x2 + y 2 + z 2 (3.1) 4λ (8µ + 7), (λ = 0, 1, 2, · · · , µ = 0, 1, 2, · · ·). (3.11) is that it should be of the form Similarly the necessary and sufficient conditions that a number cannot be expressed in the forms x2 + y 2 + 2z 2 , (3.2) x2 + y 2 + 3z 2 , (3.3) x2 + 2y 2 + 2z 2 , (3.4) x2 + 2y 2 + 3z 2 , (3.5) x2 + 2y 2 + 4z 2 , (3.6) x2 + 2y 2 + 5z 2 , (3.7) 4λ (16µ + 14), (3.21) 9λ (9µ + 6), (3.31) 4λ (8µ + 7), (3.41) 4λ (16µ + 10), (3.51) 4λ (16µ + 14), (3.61) 25λ (25µ + 10) or 25λ (25µ + 15).∗ (3.71) are that it should be of the forms Results (3.11)–(3.71) may tempt us to suppose that there are similar simple results for the form ax2 + by + cz 2 , whatever are the values of a, b, c. It appears, however, that in most cases there are no such simple results. For instance, the numbers which are not of the form x2 + 2y 2 + 10z 2 are those belonging to one or other of the four classes ∗ 2 25λ (8µ + 7), 25λ (25µ + 5), 25λ (25µ + 15), 25λ (25µ + 20). Here some of the numbers of the first class belong also to one of the next three classes. Again, the even numbers which are not of the form x2 + y 2 + 10z 2 are the numbers 4λ (16µ + 6), while the odd numbers that are not of the form, viz. 3, 7, 21, 31, 33, 43, 67, 79, 87, 133, 217, 219, 223, 253, 307, 391, · · · do not seem to obey any simple law. On the expression of a number in the form ax2 + by 2 + cz 2 + du2 221 The result concerning x2 + y 2 + z 2 is due to Cauchy: for a proof see. Landau, Handbuch der Lehre von der Verteilung der Primzahlen, p.550. The other results can be proved in an analogous manner. The form x2 + y 2 + 2z 2 has been considered by Lebesgue, and the form x2 + y 2 + 3z 2 by Dirichlet. For references see Bachmann, Zahlentheorie, Vol,IV, p.149. 4. We proceed to consider the seven cases (2.41)–(2.47). In the first case we have to shew that any number N can be expressed in the form N = x2 + y 2 + z 2 + du2 , (4.1) d being any integer between 1 and 7 inclusive. If N is not of the form 4λ (8µ + 7), we can satisfy (4.1) with u = 0. We may therefore suppose that N = 4λ (8µ + 7). First, suppose that d has one of the values 1,2,4,5,6. Take u = 2λ . Then the number N − du2 = 4λ (8µ + 7 − d) is plainly not of the form 4λ (8µ + 7), and is therefore expressible in the form x2 + y 2 + z 2 . Next, let d = 3. If µ = 0, take u = 2λ . Then N − du2 = 4λ+1 . If µ ≥ 1, take u = 2λ+1 . Then N − du2 = 4λ (8µ − 5). I have succeeded in finding a law in the following six simple cases: x2 + y 2 + 4z 2 , x2 + y 2 + 5z 2 , x2 + y 2 + 6z 2 , x2 + y 2 + 8z 2 , x2 + 2y 2 + 6z 2 , x2 + 2y 2 + 8z 2 . The numbers which are not of these forms are the numbers 4λ (8µ + 7) or 8µ + 3, 4λ (8µ + 3), 9λ (9µ + 3), 4λ (16µ + 14), 16µ + 6, or 4µ + 3, 4λ (8µ + 5), 4λ (8µ + 7) or 8µ + 5. Paper 20 222 In neither of these cases is N − du2 of the form 4λ (8µ + 7), and therefore in either case it can be expressed in the form x2 + y 2 + z 2 . Finally, let d = 7. If µ is equal to 0,1, or 2, take u = 2λ . Then N − du2 is equal to 0, 2 · 4λ+1 , or 4λ+2 . If µ ≥ 3, take u = 2λ+1 . Then N − du2 = 4λ (8µ − 21). Therefore in either case N − du2 can be expressed in the form x2 + y 2 + z 2 . Thus in all cases N is expressible in the form (4.1). Similarly we can dispose of the remaining cases, with the help of the results stated in § 3. Thus in discussing (2.42) we use the theorem that every number not of the form (3.21) can be expressed in the form (3.2). The proofs differ only in detail, and it is not worth while to state them at length. 5. We have seen that all integers without any exception can be expressed in the form m(x2 + y 2 + z 2 ) + nu2 , (5.1) when m = 1, 1 ≤ n ≤ 7, and m = 2, n = 1. We shall now consider the values of m and n for which all integers with a finite number of exceptions can be expressed in the form (5.1). In the first place m must be 1 or 2. For, if m > 2, we can choose an integer ν so that nu2 6≡ ν(mod m) for all values of u. Then (mµ + ν) − nu2 , m where µ is any positive integer, is not an integer; and so mµ + ν can certainly not be expressed in the form (5.1). We have therefore only to consider the two cases in which m is 1 or 2. First let us consider the form x2 + y 2 + z 2 + nu2 . (5.2) I shall shew that, when n has any of the values 1, 4, 9, 17, 25, 36, 68, 100, (5.21) 4k + 2, 4k + 3, 8k + 5, 16k + 12, 32k + 20, (5.22) or is of any of the forms On the expression of a number in the form ax2 + by 2 + cz 2 + du2 then all integers save a finite number, and in fact all integers from 4n onwards at any rate, can be expressed in the form (5.2); but that for the remaining values of n there is an infinity of integers which cannot be expressed in the form required. In proving the first result we need obviously only consider numbers of the form 4λ (8µ + 7) greater than n, since otherwise we may take u = 0. The numbers of this form less than n are plainly among the exceptions. 6. I shall consider the various cases which may arise in order of simplicity. (6.1) n ≡ 0(mod 8). There are an infinity of exceptions. For suppose that N = 8µ + 7. Then the number N − nu2 ≡ 7(mod 8) cannot be expressed in the form x2 + y 2 + z 2 . (6.2) n ≡ 2(mod 4). There is only a finite number of exceptions. In proving this we may suppose that N = 4λ (8µ + 7). Take u = 1. Then the number N − nu2 = 4λ (8µ + 7) − n is congruent to 1, 2, 5 or 6 to modulus 8, and so can be expressed in the form x2 + y 2 + z 2 . Hence the only numbers which cannot be expressed in the form (5.2) in this case are the numbers of the form 4λ (8µ + 7) not exceeding n. (6.3) n ≡ 5(mod 8). There is only a finite number of exceptions. We may suppose again that N = 4λ (8µ + 7). First, let λ 6= 1. Take u = 1. Then N − nu2 = 4λ (8µ + 7) − n ≡ 2 or 3(mod 8). If λ = 1 we cannot take u = 1, since N − n ≡ 7(mod 8); so we take u = 2. Then N − nu2 = 4λ (8µ + 7) − 4n ≡ 8(mod 32). In either of these cases N − nu2 is of the form x2 + y 2 + z 2 . 223 Paper 20 224 Hence the only numbers which cannot be expressed in the form (5.2) are those of the form 4λ (8µ + 7) not exceeding n, and those of the form 4(8µ + 7) lying between n and 4n. (6.4) n ≡ 3(mod 4). There is only a finite number of exceptions. Take N = 4λ (8µ + 7). if λ ≥ 1, take u = 1. Then N − nu2 ≡ 1 or 5(mod 8). if λ = 0, take u = 2. Then N − nu2 ≡ 3(mod 8). In either case the proof is completed as before. In order to determine precisely which are the exceptional numbers, we must consider more particularly the numbers between n and 4n for which λ = 0. For these u must be 1, and N − nu2 ≡ 0(mod 4). But the numbers which are multiples of 4 and which cannot be expressed in the form x2 + y 2 + z 2 are the numbers 4κ (8ν + 7), (κ = 1, 2, 3, . . . , ν = 0, 1, 2, 3, . . .). The exceptions required are therefore those of the numbers n + 4κ (8ν + 7) (6.41) which lie between n and 4n and are of the form 8µ + 7 (6.42). Now in order that (6.41) may be of the form (6.42), κ must be 1 if n is of the form 8k + 3, and κ may have any of the values 2, 3, 4, . . . if n is of the form 8k + 7. Thus the only numbers which cannot be expressed in the form (5.2), in this case, are those of the form 4λ (8µ + 7) less than n and those of the form n + 4κ (8ν + 7), (ν = 0, 1, 2, 3, · · ·), lying between n and 4n, where κ = 1 if n is of the form 8k + 3, and κ > 1 if n is of the form 8k + 7. (6.5) n ≡ 1(mod 8). In this case we have to prove that (i) if n ≥ 33, there is an infinity of integers which cannot be expressed in the form (5.2); On the expression of a number in the form ax2 + by 2 + cz 2 + du2 (ii) if n is 1, 9, 17, or 25, there is only a finite number of exceptions. In order to prove (i) suppose that N = 7.4λ . Then obviously u cannot be zero. But if u is not zero u2 is always of the form 4κ (8ν + 1). Hence N − nu2 = 7 · 4λ − n · 4κ (8ν + 1). Since n ≥ 33, λ must be greater than or equal to κ + 2, to ensure that the right-hand side shall not be negative. Hence N − nu2 = 4κ (8k + 7), where 1 k = 14 · 4λ−κ−2 − nν − (n + 7) 8 2 2 is an integer; and so N − nu is not of the form x + y 2 + z 2 . In order to prove (ii) we may suppose, as usual, that N = 4λ (8µ + 7). If λ = 0, take u = 1. Then N − nu2 = 8µ + 7 − n ≡ 6 (mod 8). If λ ≥ 1, take u = 2λ−1 . Then N − nu2 = 4λ−1 (8k + 3), where 1 k = 4(µ + 1) − (n + 7). 8 In either case the proof may be completed as before. Thus the only numbers which cannot be expressed in the form (5.2), in this case, are those of the form 8µ + 7 not exceeding n. In other words, there is no exception when n = 1; 7 is the only exception when n = 9; 7 and 15 are the only exceptions when n = 17; 7, 15 and 23 are the only exceptions when n = 25. (6.6) n ≡ 4(mod 32). By arguments similar to those used in (6.5), we can shew that (i) if n ≥ 132, there is an infinity of integers which cannot be expressed in the form (5.2); (ii) if n is equal to 4, 36, 68, or 100, there is only a finite number of exceptions, namely the numbers of the form 4λ (8µ + 7) not exceeding n. (6.7) n ≡ 20(mod 32). By arguments similar to those used in (6.3), we can shew that the only numbers which cannot be expressed in the form (5.2) are those of the form 4λ (8µ + 7) not exceeding n, and those of the form 42 (8µ + 7) lying between n and 4n. (6.8) n ≡ 12(mod 16). 225 Paper 20 226 By arguments similar to those used in (6.4), we can shew that the only numbers which cannot be expressed in the form (5.2) are those of the form 4λ (8µ + 7) less than n, and those of the form n + 4κ (8ν + 7), (ν = 0, 1, 2, 3, . . .), lying between n and 4n where κ = 2 if n is of the form 4(8k + 3) and κ > 2 if n is of the form 4(8k + 7). We have thus completed the discussion of the form (5.2), and determined the exceptional values of N precisely whenever they are finite in number. 7. We shall proceed to consider the form 2(x2 + y 2 + z 2 ) + nu2 . (7.1) In the first place n must be odd; otherwise the odd numbers cannot be expressed in this form. Suppose then that n is odd. I shall shew that all integers save a finite number can be expressed in the form (7.1); and that the numbers which cannot be so expressed are (i) the odd numbers less than n, (ii) the numbers of the form 4λ (16µ + 14) less that 4n, (iii) the numbers of the form n + 4λ (16µ + 14) greater than n less than 9n, (iv) the numbers of the form cn + 4κ (16ν + 14), (ν = 0, 1, 2, 3, . . .), greater than 9n and less than 25n, where c = 1 if n ≡ 1 (mod 4), c = 9 if n ≡ 3 (mod 4), κ = 2 if n2 ≡ 1 (mod 16), and κ > 2 if n2 ≡ 9 (mod 16). First, let us suppose N even. Then, since n is odd and N is even, it is clear that u must be even. Suppose then that u = 2v, N = 2M. We have to shew that M can be expressed in the form x2 + y 2 + z 2 + 2nv 2 . (7.2) Since 2n ≡ 2 (mod 4), it follows from (6.2) that all integers except those which are less than 2n and of the form 4λ (8µ + 7) can be expressed in the form (7.2). Hence the only even integers which cannot be expressed in the form (7.1) are those of the form 4λ (16µ + 14) less than 4n. This completes the discussion of the case in which N is even. If N is odd the discussion is more difficult. In the first place, all odd numbers less than n are plainly among the exceptions. Secondly, since n and N are both odd, u must also be odd. We can therefore suppose that N = n + 2M, u2 = 1 + 8∆, On the expression of a number in the form ax2 + by 2 + cz 2 + du2 227 where ∆ is an integer of the form 12 k(k + 1), so that ∆ may assume the values 0, 1, 3, 6, . . . And we have to consider whether n + 2M can be expressed in the form 2(x2 + y 2 + z 2 ) + n(1 + 8∆), or M in the form x2 + y 2 + z 2 + 4n∆. (7.3) If M is not of the form 4λ (8µ + 7), we can take ∆ = 0. If it is of this form, and less than 4n, it is plainly an exception. These numbers give rise to the exceptions specified in (iii) of section 7. We may therefore suppose that M is of the form 4λ (8µ + 7) and greater than 4n. 8. In order to complete the discussion, we must consider the three cases in which n ≡ 1 (mod 8), n ≡ 5 (mod 8), and n ≡ 3 (mod 4) separately. (8.1) n ≡ 1 mod 8). If λ is equal to 0,1, or 2, take ∆ = 1. Then M − 4n∆ = 4λ (8µ + 7) − 4n is of one of the forms 8ν + 3, 4(8ν + 3), 4(8ν + 6). If λ ≥ 3 we cannot take ∆ = 1, since M − 4n∆ assumes the form 4(8ν + 7); so we take ∆ = 3. Then M − 4n = ∆4λ (8µ + 7) − 12n is of the form 4(8ν + 5). In either of these cases M − 4n∆ is of the form x2 + y 2 + z 2 . Hence the only values of M , other than those already specified which cannot be expressed in the form (7.3), are those of the form 4κ (8ν + 7), (ν = 0, 1, 2, . . . , κ > 2), lying between 4n and 12n. In other words, the only numbers greater than 9n which cannot be expressed in the form (7.1), in this case, are the numbers of the form n + 4κ (8ν + 7), (ν = 0, 1, 2, . . . , κ > 2), lying between 9n and 25n. (8.2) n ≡ 5 mod 8). if λ 6= 2, take ∆ = 1. Then M − 4n∆ = 4λ (8µ + 7) − 4n is of one of the forms 8ν + 3, 4(8ν + 2), 4(8ν + 3). Paper 20 228 If λ = 2, we cannot take ∆ = 1, since M − 4n∆ assumes the form 4(8ν + 7); so we take ∆ = 3. Then M − 4n∆ = 4λ (8µ + 7) − 12n is of the form 4(8ν + 5). In either of these cases M − 4n∆ is of the form x2 + y 2 + z 2 . Hence the only values of M , other than those already specified, which cannot be expressed in the form (7.3), are those of the form 16(8µ + 7) lying between 4n and 12n. In other words, the only numbers greater than 9n which cannot be expressed in the form (7.1), in this case, are the numbers of the form n + 42 (16µ + 14) lying between 9n and 25n. (8.3) n ≡ 3 (mod 4). If λ 6= 1, take ∆ = 1. Then M − 4n∆ = 4λ (8µ + 7) − 4n is of one of the forms 8ν + 3, 4(4ν + 1). If λ = 1, take ∆ = 3. Then M − 4n∆ = 4(8µ + 7) − 12n is of the form 4(4ν + 2). In either of these cases M − 4n∆ is of the form x2 + y 2 + z 2 . This completes the proof that there is only a finite number of exceptions. In order to determine what they are in this case, we have to consider the values of M , between 4n and 12n, for which ∆ = 1 and M − 4n∆ = 4(8µ + 7 − n) ≡ 0 mod 16). But the numbers which are multiples of 16 and which cannot be expressed in the form x2 + y 2 + z 2 are the numbers 4κ (8ν + 7), (κ = 2, 3, 4, . . . , ν = 0, 1, 2, . . .). The exceptional values of M required are therefore those of the numbers 4n + 4κ (8ν + 7) (8.31) which lie between 4n and 12n and are of the form 4(8µ + 7). (8.32) But in order that (8.31) may be of the form (8.32), κ must be 2 if n is of the form 8k + 3, and κ may have any of the values 3, 4, 5, . . . if n is of the form 8k + 7. It follows that the only numbers greater than 9n which cannot be expressed in the form (7.1), in this case, are the numbers of the form 9n + 4κ (16ν + 14), (ν = 0, 1, 2, . . .), lying between 9n and 25n, where κ = 2 if n is of the form 8k + 3, and κ > 2 if n is of the form 8k + 7. This completes the proof of the results stated in section 7.